max
0
s
.
t
.
a
1
x
1
+
a
2
x
2
≤
b
x
1
, x
2
≥
0
.
Lizhi Wang ([email protected])
IE 534 Linear Programming
October 31, 2012
8 / 21
Totally unimodular matrix
If the
A
matrix has a special property (totally unimodular) and
b
is
integral, then
(
A, b
) = (
˜
A,
˜
b).A matrix istotally unimodular(TU) if the determinant of everysquare submatrix is0,1, or1.IA subset of columns and rows of a matrix is a submatrix.IA11A14A31A34is a submatrix ofA11A12A13A14A21A22A23A24A31A32A33A34IfAis TU, thenAIis also TU. (There are many similarproperties).The nodearc incidence matrix of a directed graph is TU.
Lizhi Wang ([email protected])
IE 534 Linear Programming
October 31, 2012
9 / 21
Example

4
6
1
1
is not TU, because
det

4
6
1
1
=

10
,
det [

4] =

4
, and
det [6] = 6
.
1

1
0
0
0
0
0
0

1
0
1

1
0
1
0
0
0
1

1
1
1
0
1
0
0
0
0
0

1

1
0
1
0
0
0
0
0
0

1

1
is TU, because it’s a
nodearc incidence matrix.
Lizhi Wang ([email protected])
IE 534 Linear Programming
October 31, 2012
10 / 21
Nodearc incidence matrix
Suppose a graph has
n
nodes and
m
arcs
The nodearc incidence
matrix is an
n
×
m
matrix
For each arc
k
= (
i, j
)
,
A
ik
= 1
and
A
jk
=

1
Other elements are zeros

1

2

3

4

5

6

7

8
A
B
C
D
E
F
1
1
0
0
0
0
0

1

1
0
1
0
0
0
0
1
0

1

1

1
1
1
0
0
0
0
0
1
0
0

1
0
0
0
0
0
0

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 Spring '12
 lizhiwang
 Linear Programming, Vector Space, Optimization, Linear Programming Relaxation, Constraint, Lizhi Wang